12 Examples of Subsets that Are Not Subspaces of Vector Spaces

Problem 338

Each of the following sets are not a subspace of the specified vector space. For each set, give a reason why it is not a subspace.(1) \[S_1=\left \{\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} \in \R^3 \quad \middle | \quad x_1\geq 0 \,\right \}\]
in the vector space $\R^3$.

Solution.

Recall the following subspace criteria.
A subset $W$ of a vector space $V$ over the scalar field $K$ is a subspace of $V$ if and only if the following three criteria are met.

The subset $W$ contains the zero vector of $V$.

If $u, v\in W$, then $u+v\in W$.

If $u\in W$ and $a\in K$, then $au\in W$.

Thus, to prove a subset $W$ is not a subspace, we just need to find a counterexample of any of the three criteria.

Solution (1). $S_1=\{ \mathbf{x} \in \R^3 \mid x_1\geq 0 \}$

The subset $S_1$ does not satisfy condition 3. For example, consider the vector
\[\mathbf{x}=\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}.\]
Then since $x_1=1\geq 0$, the vector $\mathbf{x}\in S_1$. Then consider the scalar product of $\mathbf{x}$ and the scalar $-1$. Then we have
\[(-1)\cdot\mathbf{x}=\begin{bmatrix}
-1 \\
0 \\
0
\end{bmatrix},\]
and the first entry is $-1$, hence $-\mathbf{x}$ is not in $S_1$. Thus $S_1$ does not satisfy condition 3 and it is not a subspace of $\R^3$.
(You can check that conditions 1, 2 are met.)

Solution (2). $S_2= \{ \mathbf{x}\in \R^3\mid x_1-4x_2+5x_3=2 \}$

The zero vector of the vector space $\R^3$ is
\[\mathbf{0}=\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}.\]
Since the zero vector $\mathbf{0}$ does not satisfy the defining relation $x_1-4x_2+5x_3=2$, it is not in $S_2$. Hence condition 1 is not met, hence $S_2$ is not a subspace of $\R^3$.
(You can check that conditions 2, 3 are not met as well.)

Solution (3). $S_3=\{\mathbf{x}\in \R^2 \mid y=x^2 \quad \}$

However, their sum
\[\begin{bmatrix}
1 \\
1
\end{bmatrix} + \begin{bmatrix}
-1 \\
1
\end{bmatrix}
=
\begin{bmatrix}
0 \\
1
\end{bmatrix}\]
is not in $S_3$ since $1\neq 0^2$.
Hence condition 2 is not met, and thus $S_3$ is not a subspace of $\R^2$.
(You can check that condition 1 is fulfilled yet condition 3 is not met.)

Solution (4). $S_4=\{ f(x)\in P_4 \mid f(1) \text{ is an integer}\}$

Consider the polynomial $f(x)=x$. Since the degree of $f(x)$ is $1$ and $f(1)=1$ is an integer, it is in $S_4$. Consider the scalar product of $f(x)$ and the scalar $1/2\in \R$.
Then we evaluate the scalar product at $x=1$ and we have
\begin{align*}
\frac{1}{2}f(1)=\frac{1}{2},
\end{align*}
which is not an integer.
Thus $(1/2)f(x)$ is not in $S_4$, hence condition 3 is not met. Thus $S_4$ is not a subspace of $P_4$.
(You can check that conditions 1, 2 are met.)

Let $f(x)=x$. Then $f(x)$ is a degree $1$ polynomial and $f(1)=1$ is a rational number.
However, the scalar product $\sqrt{2} f(x)$ of $f(x)$ and the scalar $\sqrt{2} \in \R$ is not in $S_5$ since
\[\sqrt{2}f(1)=\sqrt{2},\]
which is not a rational number. Hence condition 3 is not met and $S_5$ is not a subspace of $P_4$.
(You can check that conditions 1, 2 are met.)

Solution (6). $S_6=\{ A\in M_{2\times 2} \mid \det(A) \neq 0\}$

The zero vector of the vector space $M_{2 \times 2}$ is the $2\times 2$ zero matrix $O$.
Since the determinant of the zero matrix $O$ is $0$, it is not in $S_6$. Thus, condition 1 is not met and $S_6$ is not a subspace of $M_{2 \times 2}$.
(You can check that conditions 2, 3 are not met as well.)

Solution (7). $S_7=\{ A\in M_{2\times 2} \mid \det(A)=0\}$

Consider the matrices
\[A=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix} \text{ and } B=\begin{bmatrix}
0 & 0\\
0& 1
\end{bmatrix}.\]
The determinants of $A$ and $B$ are both $0$, hence they belong to $S_7$.
However, their sum
\[A+B=\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}\]
has the determinant $1$, hence the sum $A+B$ is not in $S_7$.
So condition 2 is not met and $S_7$ is not a subspace of $M_{2 \times 2}$.
(You can check that conditions 1, 3 are met.)

Solution (8). $S_8=\{ f(x)\in C[-2,2] \mid f(-1)f(1)=0\}$

Consider the continuous functions
\[f(x)=x-1 \text{ and } g(x)=x+1.\]
(These are polynomials, hence they are continuous.)
We have
\begin{align*}
&f(-1)f(1)=(-2)\cdot(0)=0 \text{ and }\\
&g(-1)g(1)=(0)\cdot 2=0.
\end{align*}
So these functions are in $S_8$.

However, their sum $h(x):=f(x)+g(x)$ does not belong to $S_8$ since we have
\begin{align*}
h(-1)h(1)&=\big(f(-1)+g(-1)\big) \big(f(1)+g(1) \big)\\
&=(-2+0)(0+2)=-4\neq 0.
\end{align*}
Therefore, condition 2 is not met and $S_8$ is not a subspace of $C[-1, 1]$.
(You can check that conditions 1, 3 are met.)

Let $f(x)=x^2$, an open-up parabola.
Then $f(x)$ is continuous and non-negative for $-1 \leq x \leq 1$. Hence $f(x)=x^2$ is in $S_9$.
However, the scalar product $(-1)f(x)$ of $f(x)$ and the scalar $-1$ is not in $S_9$ since, say,
\[(-1)f(1)=-1\]
is negative.
So condition 3 is not met and $S_9$ is not a subspace of $C[-1, 1]$.
(You can check that conditions 1, 2 are met.)

The zero vector of the vector space $C^2[-1, 1]$ is the zero function $\theta(x)=0$.
The second derivative of the zero function is still the zero function.
Thus,
\[\theta^{\prime\prime}(x)+\theta(x)=0\]
and since $\sin(x)$ is not the zero function, $\theta(x)$ is not in $S_{10}$.
Hence $S_{10}$ is not a subspace of $C^2[-1, 1]$.

(You can check that conditions 2, 3 are not met as well.
For example, consider the function $f(x)=-\frac{1}{2}x\cos(x)\in S_{10}$.)

Solution (11). Let $S_{11}$ be the set of real polynomials of degree exactly $k$.

The set $S_{11}$ is not a vector subspace of $\mathbf{P}_k$. One reason is that the zero function $\mathbf{0}$ has degree $0$, and so does not lie in $S_{11}$. The set $S_{11}$ is also not closed under addition. Consider the two polynomials $f(x) = x^k + 1$ and $g(x) = – x^k + 1$. Both of these polynomials lie in $S_{11}$, however $f(x) + g(x) = 2$ has degree $0$ and so does not lie in $S_{11}$.

Solution (12). The complement

The complement $S_{12}= V \setminus W$ is not a vector subspace. Specifically, if $\mathbf{0} \in V$ is the zero vector, then we know $\mathbf{0} \in W$ because $W$ is a subspace. But then $\mathbf{0} \not\in V \setminus W$, and so $V \setminus W$ cannot be a vector subspace.

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